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Use Definition 2 to find an expression for the area under the graph of $ f $ as a limit. Do not evaluate the limit.
$ f(x) = \dfrac{2x}{x^2 + 1}, \hspace{5mm} 1 \le x \le 3 $
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02:08
Frank Lin
00:44
Amrita Bhasin
Calculus 1 / AB
Chapter 5
Integrals
Section 1
Areas and Distances
Integration
Dishant P.
September 30, 2020
hui
Campbell University
Harvey Mudd College
Baylor University
University of Nottingham
Lectures
05:53
In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.
40:35
In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.
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Use Definition 2 to find a…
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$$\begin{array}{l}{\text {…
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17-19 Use Definition 2 to …
So the general theme when we start talking about Integral is the idea of area under the curve on definition two states that the area is equal to the limit as an approaches infinity of Delta X, the change in X times The summation from I equals one toe end of f of x abi. So in this case, we know that Delta X is equal to B minus a our ending point in R minus our initial point divided by the number of kind of rectangles we have or whatever we're using to split up the area. So what this problem that we have is f of X being equal to two X over X squared, plus one. So we want to set up our limit. Um, we know that a equals one and B equals three. So right off the bat, we confined Delta X to be equal to three minus one over n. So it'll be too over end. Then, um, with this information, we see that the area will be equal to the limit as an approaches infinity of two over em times the sum from I equals one toe end of two times our X value and we know they're X value is one plus Chu I over end over X squared. So are once again or X values me one plus two i over and squared plus one Andi. The way that we can get that is knowing that we're evaluating X being equal to one plus I times two over end. Once we plug that into ffx, we'll be able to do everything that we see right here for our final answer.
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